# Compound Events Examples

The subsets of the sample space is known as events. The events having more than one occurrences happening together are known as compound events. The sample space of compound events can be represented by tree diagram, lists and table. If two coins are tossed simultaneously then it will be called a compound event as more than one occurrence is involved in the experiment. The sample can be listed as, $\{(H, H),\ (H, T),\ (T, H),\ (T, T)\}$. The probability of compound event $E$ is given by the ratio of number of favorable events to total number of events in the sample space.

## Word Problem

Problem 1:

Find the sample space of tossing a coin and choosing a ball out of four balls, colored green, blue, yellow, and red using a table.

Solution:

Given is the table to show the sample space.

 Head Tail Green G-H G-T Blue B-H B-T Yellow Y-H Y-T Red R-H R-T
Problem 2:

There are three books of maths, geography and economics who are to be placed on shelves 1, 2 and 3 randomly. In how many ways they can be arranged?

Solution:

The given tree diagram will represent the ways these books can be arranged.

Problem 3:

A coin is tossed and one card is chosen from four cards where all four are queens. Find the sample space in the listed form.

Solution:

Let four queen cards be $Q1,\ Q2,\ Q3,$ and $Q4$. The tossing of coin will give two outcomes as head and tail.

The sample space is listed here.

$\{(Q1,\ H),\ (Q1,\ T),\ (Q2,\ H),\ (Q2,\ T),\ (Q3,\ H),\ (Q3,\ T),\ (Q4,\ H),\ (Q4,\ T)\}$
Problem 4:

Find the sample space if four coins are tossed together.

Solution:

Given is the tree diagram for the required sample space.

Problem 5:

If a coin is tossed and a marble is chosen from red, white and green marbles, then find the probability of getting a red marble.

Solution:

The sample space is given by,

 H T Green G-H G-T White W-H W-T Red R-H R-T

Number of events in sample space = $n(S)$ = $6$

Number of favorable events, that is, events having red marble = $n(E)$ = $2$

Probability of getting red marble, $P(E)$ = $\frac{n(E)}{n(S)}$ = $\frac{2}{6}$ = $\frac{1}{3}$